How do you account for the reducing behaviour of $H_{3}PO_{2}$ on the basis of its structure?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The structure of $H_{3}PO_{2}$ (hypophosphorous acid) contains two $P-H$ bonds and one $P=O$ bond,along with one $P-OH$ bond.
Because of the presence of two $P-H$ bonds,$H_{3}PO_{2}$ acts as a strong reducing agent.
The $H$ atoms bonded directly to the $P$ atom are responsible for its reducing character.

Explore More

Similar Questions

The reaction of aqueous $NaOH$ with white $P_4$ is given by: $P_4 + NaOH (aq) \rightarrow PH_3 + X$. The compound $X$ is:

$P_4 + 3NaOH + 3H_2O \to A + 3NaH_2PO_2$. Here,$A$ is:

White phosphorus contains

Extra pure $N_2$ can be obtained by heating :-

An orange solid $(X)$ on heating gives a colourless gas $(Y)$ and only a green residue $(Z)$. Gas $(Y)$ on treatment with $Mg$ produces a white solid substance. Identify the white solid substance.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo